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Chemical Kinetics - Integrated rate equations (Zero and First order)

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Zero Order Reactions: These are reactions where the rate is independent of the concentration of reactants. The rate law is expressed as Rate=k[A]0=kRate = k[A]^0 = k. The concentration of reactants decreases linearly with time.

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First Order Reactions: In these reactions, the rate is directly proportional to the concentration of one reactant. The rate law is expressed as Rate=k[A]1Rate = k[A]^1. Decomposition of radioactive nuclei is a classic example of first-order kinetics.

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Integrated Rate Equations: These equations relate the concentration of reactants with time, allowing for the calculation of rate constants and the prediction of concentrations at any given time tt.

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Half-life (t1/2t_{1/2}): For a zero-order reaction, the half-life is directly proportional to the initial concentration (t1/2∝[R]0t_{1/2} \propto [R]_0). For a first-order reaction, the half-life is independent of the initial concentration (t1/2=0.693kt_{1/2} = \frac{0.693}{k}).

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Graphical Representation: For zero order, a plot of [R][R] vs tt gives a straight line with slope −k-k. For first order, a plot of log⁡[R]\log [R] vs tt gives a straight line with slope −k2.303-\frac{k}{2.303}.

📐Formulae

[R]t=−kt+[R]0 (Zero Order Integrated Equation)[R]_t = -kt + [R]_0 \text{ (Zero Order Integrated Equation)}

t1/2=[R]02k (Zero Order Half-life)t_{1/2} = \frac{[R]_0}{2k} \text{ (Zero Order Half-life)}

k=2.303tlog⁡[R]0[R]t (First Order Integrated Equation)k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t} \text{ (First Order Integrated Equation)}

t1/2=0.693k (First Order Half-life)t_{1/2} = \frac{0.693}{k} \text{ (First Order Half-life)}

[R]t=[R]0e−kt (Exponential form for First Order)[R]_t = [R]_0 e^{-kt} \text{ (Exponential form for First Order)}

💡Examples

Problem 1:

A first-order reaction is found to have a rate constant k=5.5×10−14 s−1k = 5.5 \times 10^{-14} \text{ s}^{-1}. Find the half-life of the reaction.

Solution:

t1/2=0.693k=0.6935.5×10−14 s−1≈1.26×1013 st_{1/2} = \frac{0.693}{k} = \frac{0.693}{5.5 \times 10^{-14} \text{ s}^{-1}} \approx 1.26 \times 10^{13} \text{ s}

Explanation:

Since the reaction is first-order, the half-life is calculated using the formula t1/2=0.693kt_{1/2} = \frac{0.693}{k}, which shows the half-life is independent of the initial concentration.

Problem 2:

A first-order reaction takes 4040 minutes for 30%30\% decomposition. Calculate its t1/2t_{1/2}.

Solution:

Step 1: Find kk. Let [R]0=100[R]_0 = 100, then [R]t=100−30=70[R]_t = 100 - 30 = 70. Using k=2.303tlog⁡[R]0[R]t=2.30340log⁡10070=2.30340×0.1549=8.91×10−3 min−1k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t} = \frac{2.303}{40} \log \frac{100}{70} = \frac{2.303}{40} \times 0.1549 = 8.91 \times 10^{-3} \text{ min}^{-1}. \nStep 2: Find half-life. t1/2=0.693k=0.6938.91×10−3≈77.7 minutest_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.91 \times 10^{-3}} \approx 77.7 \text{ minutes}.

Explanation:

We first use the integrated rate law for first-order kinetics to find the rate constant kk from the given time and percentage of completion, then use that kk to find the half-life.

Integrated rate equations (Zero and First order) Revision - Class 12 Chemistry ICSE